Answer:
See below
Step-by-step explanation:
By definition the Laplace transform of f(t) = t-6 is
[tex] \bf F(s)=L[f(t)](s)=\int_{0}^{\infty}e^{-st}(t-6)dt[/tex]
To find an anti-derivative, let's consider the indefinite integral
[tex] \bf \int e^{-st}(t-6)dt[/tex]
and use integration by parts.
If we set
[tex] \bf u=(t-6)\\dv=e^{-st}dt[/tex]
then
[tex] \bf du=dt\\v=\int e^{-st}dt=\frac{e^{-st}}{-s}[/tex]
hence
[tex] \bf \int e^{-st}(t-6)dt=uv-\int vdu=(t-6)\frac{e^{-st}}{-s}-\int \frac{e^{-st}}{-s}dt=\\=(t-6)\frac{e^{-st}}{-s}+\frac{1}{s}\int e^{-st}dt=(t-6)\frac{e^{-st}}{-s}+\frac{1}{s}\frac{e^{-st}}{(-s)}=\\=(t-6)\frac{e^{-st}}{-s}-\frac{e^{-st}}{s^2}=\frac{e^{-st}}{s}(6-t-\frac{1}{s})[/tex]
To find the Laplace transform we evaluate the improper integral as
[tex] \bf \lim_{n \rightarrow \infty}\int_{0}^{n}e^{-st}(t-6)dt=\lim_{n \rightarrow \infty}\left[\frac{e^{-st}}{s}(6-t-\frac{1}{s}) \right]_0^{n}=\\=\lim_{n \rightarrow \infty}\left[\frac{e^{-sn}}{s}(6-n-\frac{1}{s}) -\frac{1}{s}(6-\frac{1}{s})\right][/tex]
But
[tex] \bf \lim_{n \rightarrow \infty}\left[\frac{e^{-sn}}{s}(6-n-\frac{1}{s})\right]=0\;(s>0)[/tex]
and does not exist for s<0.
So, the Laplace transform only exists for s>0 and equals
[tex] \bf \boxed{F(s)=-\frac{1}{s}(6-\frac{1}{s})=6(\frac{1}{s^2}-\frac{1}{s})}[/tex] for s>0
